Question

☆ Topic : Heating and Magnetic Effect of electric current
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Q. You are provided with two bulbs operating on 220 V, one of 60 A and the other of 100 W. If the two are connected in series and then the combination is connected to 440 V, which one of them will fuse and why?
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Answer 1

ANSWER:-

Given:-

The total current flowing through the series circuit will be

$\sf{I = \frac{V}{R} = \frac{440}{R} }$

Here, R is the total resistance offered by the two devices

$\tt{R = R _{1} + R_{2} = \frac{V {}^{2} _{1} }{P_{1} } + \frac{V {}^{2}_{2} }{P _{2} } = \frac{220 {}^{2} }{60} + \frac{220 {}^{2} }{100} = 806.7 + 484 = 1290.7 }$

$\rm{so \: I = \frac{440}{1290.7} = 0.34 \: A}$

The maximum current allowed by two devices will be

$\mathbb{I _{1} = \frac{P_{1} }{V_{1} } = \frac{60}{220} = 0.3 \: A}$

$\sf{I_{2} = \frac{P_{2}}{V_{2} } = \frac{100}{220} = 0.45 \: A}$

As the current supplier is greater than the rated

current for the first bulb. So it will fuse.

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